You are right when you say that $\forall x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true.
Then you say "the statement $\forall x \in \emptyset$ is false". $\forall x \in \emptyset$ is NOT a statement, it's an incomplete sentence. Either you write "$\forall x, P(x)$", either you write "$\forall x \in X, P(x)$", which is a shorthand for "$\forall x, (x \in X \implies P(x))$". "$\forall x \in \emptyset$" is not a statement. It can't be true or false.
$\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, ~\textrm{false} \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, ~\textrm{true}$ (whatever $P(x)$ may be), which is $\textrm{true}\;.$
If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$.
The first negation is almost completely right. You forgot to negate the implication at the end.
Remember that the negation of an "if, then" statement is not an "if, then" statement. $A \implies B$ has negation "$A \land \neg B$" (read: $A$ and not $B$).
So, "if the sky is blue, then I love cheese" has negation "the sky is blue and I do not love cheese."
We say $\lim \limits_{x \to a} f(x) = L$ if $$\forall \epsilon > 0\text{, }\exists \delta > 0 \text{ such that }\forall x\text{, }|x - a| < \delta \implies |f(x) - L| < \epsilon.$$ Then the negation of this is: $$\exists \epsilon > 0\text{ such that }\forall \delta > 0\text{, }\exists x\text{ such that }|x - a| < \delta \text{ **and** }|f(x) - L| \geq \epsilon.$$
UPDATE Here is how to negate the following statement step by step
Negation of "$\lim \limits_{x \to a} f(x)$ exists", i.e., $$\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| \geq \epsilon).$$
We say $\lim \limits_{x \to a} f(x)$ does not exist if:
$\neg[\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$
$\forall L \neg[\forall\epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$
$\forall L\exists \epsilon > 0\neg[\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$
$\forall L\exists \epsilon > 0\forall \delta > 0\neg[\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$
$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: \neg[|x - a| < \delta \implies |f(x) - L| < \epsilon]$
$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land \neg(|f(x) - L| < \epsilon)$ (Negation of implication)
$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$
Best Answer
Yes, your proof is correct. It follows from the formal definition of injectivity ($P$) that if there exists even one counter-example, the function is no longer injective.
$\exists \; x, y \in E, \; x \neq y: f(x) = f(y)$
equals
$\neg \forall \; x, y \in E: f(x) = f(y) \Rightarrow x = y$
Which is the negation of $P$.